Topic 16

Interval Estimation

Our strategy to estimation thus far has been to use a method to find an , e.g., method of moments, or maximum likelihood, and evaluate the quality of the estimator by evaluating the bias and the of the estimator. Often, we know more about the distribution of the estimator and this allows us to take a more comprehensive statement about the estimation procedure. is an alternative to the variety of techniques we have examined. Given x, we replace the point estimate ✓ˆ(x) for the ✓ by a that is subset Cˆ(x) of the parameter space. We will consider both the classical and Bayesian approaches to choosing Cˆ(x) . As we shall learn, the two approaches have very different interpretations.

16.1 Classical

In this case, the random set Cˆ(X) is chosen to have a prescribed high , , of containing the true parameter value ✓. In symbols, P ✓ Cˆ(X) = . ✓{ 2 } In this case, the set Cˆ(x) is called a -level con-

fidence set. In the case of a one dimensional param- 0.4 eter set, the typical choice of confidence set is a con-

fidence interval 0.35

Cˆ(x)=(✓ˆ`(x), ✓ˆu(x)). 0.3 Often this interval takes the form 0.25 Cˆ(x)=(✓ˆ(x) m(x), ✓ˆ(x)+m(x)) = ✓ˆ(x) m(x) ± where the two statistics, 0.2 ✓ˆ(x) is a point estimate, and • 0.15 m(x) is the margin of error. • 0.1

16.1.1 0.05 area α Example 16.1 (1- z interval). If X1.X2....Xn 0 µ z are normal random variables with unknown −3 −2 −1 0 1 2 α 3 2 but known variance 0. Then, . ↵ is the area under the standard X¯ µ Figure 16.1: Upper tail critical values Z = normal density and to the right of the vertical line at critical value z↵ 0/pn

239 Introduction to the Science of Statistics Interval Estimation

is a standard normal . For any ↵ between 0 and 1, let z↵ satisfy P Z>z = ↵ or equivalently P Z z =1 ↵. { ↵} {  ↵} The value is known as the upper tail probability with critical value z↵. We can compute this in R using, for example > qnorm(0.975) [1] 1.959964 for ↵ =0.025. If =1 2↵, then ↵ =(1 )/2. In this case, we have that P z

X¯ µ0 = Z

Similarly, X¯ µ0 = Z> z↵ 0/pn implies µ < X¯ + z 0 0 ↵ pn Thus X¯ z 0 <µ < X¯ + z 0 . ↵ pn 0 ↵ pn has probability . Thus, for data x, 0 x¯ z(1 )/2 ± pn is a confidence interval with confidence level . In this case,

µˆ(x)=¯x is the estimate for the mean and m(x)=z(1 )/20/pn is the margin of error. We can use the z-interval above for the confidence interval for µ for data that is not necessarily normally dis- tributed as long as the applies. For one population tests for means, n>30 and data not strongly skewed is a good rule of thumb. Generally, the is not known and must be estimated. So, let X ,X , ,X be normal random 1 2 ··· n variables with unknown mean and unknown standard deviation. Let S2 be the unbiased sample variance. If we are forced to replace the unknown variance 2 with its unbiased estimate s2, then the statistic is known as t: x¯ µ t = . s/pn

The term s/pn which estimates the standard deviation of the sample mean is called the . The remarkable discovery by William Gossett is that the distribution of the t statistic can be determined exactly. Write pn(X¯ µ) Tn 1 = . S Then, Gossett was able to establish the following three facts:

240 Introduction to the Science of Statistics Interval Estimation 0.4 1.0 0.8 0.3 0.6 0.2 0.4 0.1 0.2 0.0 0.0

-4 -2 0 2 4 -4 -2 0 2 4

x x

Figure 16.2: The density and distribution function for a standard normal random variable (black) and a t random variable with 4 degrees of freedom (red). The variance of the t distribution is df / (df 2) = 4/(4 2) = 2 is higher than the variance of a standard normal. This can be seen in the broader shoulders of the t density function or in the smaller increases in the t distribution function away from the mean of 0.

The numerator is a standard normal random variable. • The denominator is the square root of • 1 n S2 = (X X¯)2. n 1 i i=1 X The sum has chi-square distribution with n 1 degrees of freedom. The numerator and denominator are independent. • With this, Gossett was able to compute the density of the t distribution with n 1 degrees of freedom. Gossett, who worked for the the brewery of Arthur Guinness in Dublin, was permitted to publish his results only if it appeared under a pseudonym. Gosset chose the name Student, thus the distribution is sometimes known as Student’s t.

Again, for any ↵ between 0 and 1, let upper tail probability tn 1,↵ satisfy

P Tn 1 >tn 1,↵ = ↵ or equivalently P Tn 1 tn 1,↵ =1 ↵. { } {  } We can compute this in R using, for example > qt(0.975,12) [1] 2.178813 for ↵ =0.025 and n 1 = 12. Example 16.2. For the data on the lengths of 200 Bacillus subtilis, we had a mean x¯ =2.49 and standard deviation s =0.674. For a 96% confidence interval ↵ =0.02 and we type in R, > qt(0.98,199) [1] 2.067298

241 Introduction to the Science of Statistics Interval Estimation 4 3 2 1 0

10 20 30 40 50

df

Figure 16.3: Upper critical values for the t confidence interval with =0.90 (black), 0.95 (red), 0.98 (magenta) and 0.99 (blue) as a function of df , the number of degrees of freedom. Note that these critical values decrease to the critical value for the z confidence interval and increases with .

Thus, the interval is 0.674 2.490 2.0674 =2.490 0.099 or (2.391, 2.589) ± p200 ± Example 16.3. We can obtain the data for the Michaelson-Morley using R by typing > data(morley) The data have 100 rows - 5 (column 1) of 20 runs (column 2). The Speed is in column 3. The values for speed are the amounts over 299,000 km/sec. Thus, a t-confidence interval will have 99 degrees of freedom. We can see a by writing hist(morley$Speed). To determine a 95% confidence interval, we find > mean(morley$Speed) [1] 852.4 > sd(morley$Speed) [1] 79.01055 > qt(0.975,99) [1] 1.984217 Thus, our confidence interval for the speed of light is 79.0 299, 852.4 1.9842 = 299, 852.4 15.7 or the interval (299836.7, 299868.1) ± p100 ± This confidence interval does not include the presently determined values of 299,792.458 km/sec for the speed of light. The confidence interval can also be found by tying t.test(morley$Speed). We will study this command in more detail when we describe the t-test.

242 Introduction to the Science of Statistics Interval Estimation

Histogram of morley$Speed 30 25 20 15 10 5 0

600 700 800 900 1000 1100

morley$Speed

Figure 16.4: Measurements of the speed of light. Actual values are 299,000 kilometers per second plus the value shown.

Exercise 16.4. Give a 90% and a 98% confidence interval for the example above. We often wish to determine a sample size that will guarantee a desired margin of error. For a -level t-interval, this is s m = tn 1,(1 )/2 . pn Solving this for n yields 2 tn 1,(1 )/2 s n = . m ✓ ◆ Because the number of degrees of freedom, n 1, for the t distribution is unknown, the quantity n appears on both sides of the equation and the value of s is unknown. We search for a conservative value for n, i.e., a margin of error that will be no greater that the desired length. This can be achieved by overestimating tn 1,(1 )/2 and s. For the speed of light example above, if we desire a margin of error of m = 10 km/sec for a 95% confidence interval, then we set tn 1,(1 )/2 =2and s = 80 to obtain 2 80 2 n · = 256 ⇡ 10 ✓ ◆ measurements are necessary to obtain the desired margin of error.. The next set of confidence intervals are determined, in the case in which the distributional variance in known, by finding the standardized score and using the normal approximation as given via the central limit theorem. In the cases in which the variance is unknown, we replace the distribution variance with a variance that is estimated from the observations. In this case, the procedure that is analogous to the standardized score is called the studentized score. Example 16.5 (matched pair t interval). We begin with two quantitative measurements

(X1,1,...,X1,n) and (X2,1,...,X2,n), on the same n individuals. Assume that the first set of measurements has mean µ1 and the second set has mean µ2.

243 Introduction to the Science of Statistics Interval Estimation

If we want to determine a confidence interval for the difference µ µ , we can apply the t-procedure to the 1 2 differences (X X ,...,X X ) 1,1 2,1 1,n 2,n to obtain the confidence interval ¯ ¯ Sd (X1 X2) tn 1,(1 )/2 ± pn where Sd is the standard deviation of the difference.

Example 16.6 (2-sample z interval). If we have two independent samples of normal random variables

(X1,1,...,X1,n1 ) and (X2,1,...,X2,n2 ),

2 2 the first having mean µ1 and variance 1 and the second having mean µ2 and variance 2, then the difference in their sample means X¯ X¯ 2 1 is also a normal random variable with

2 2 1 2 mean µ1 µ2 and variance + . n1 n2 Therefore, (X¯ X¯ ) (µ µ ) Z = 1 2 1 2 2 2 1 + 2 n1 n2 q 2 2 is a standard normal random variable. In the case in which the 1 and 2 are known, this gives us a -level confidence interval for the difference in µ µ . 1 2

2 2 ¯ ¯ 1 2 (X1 X2) z(1 )/2 + . ± sn1 n2

2 2 Example 16.7 (2-sample t interval). If we know that 1 = 2, then we can pool the data to compute the standard 2 2 deviation. Let S1 and S2 be the sample variances from the two samples. Then the pooled sample variance Sp is the weighted average of the sample variances with weights equal to their respective degrees of freedom.

(n 1)S2 +(n 1)S2 S2 = 1 1 2 2 . p n + n 2 1 2 This give a statistic (X¯1 X¯2) (µ1 µ2) Tn1+n2 2 = S 1 + 1 p n1 n2 that has a t distribution with n + n 2 degrees of freedom.q Thus we have the level confidence interval 1 2 ¯ ¯ 1 1 (X1 X2) tn +n 2,(1 )/2Sp + ± 1 2 n n r 1 2 for µ1 µ2. 2 2 If we do not know that 1 = 2, then the corresponding studentized random variable

(X¯ X¯ ) (µ µ ) T = 1 2 1 2 S2 S2 1 + 2 n1 n2 q 244 Introduction to the Science of Statistics Interval Estimation no longer has a t-distribution. Welch and Satterthwaite have provided an approximation to the t distribution with effective degrees of freedom given by the Welch-Satterthwaite equation

2 2 2 s1 + s2 n1 n2 ⌫ = 4 4 . (16.1) s1⇣ ⌘s2 n2 (n 1) + n2 (n 1) 1· 1 2· 2 This give a -level confidence interval

2 2 s1 s2 x¯1 x¯2 t⌫,(1 /2 + . ± sn1 n2

For two sample tests, the number of observations per group may need to be at least 40 for a good approximation to the .

Exercise 16.8. Show that the effective degrees is between the worst case of the minimum choice from a one sample t-interval and the best case of equal variances.

min n ,n 1 ⌫ n + n 2 { 1 2}   1 2 For data on the life span in days of 88 wildtype and 99 transgenic mosquitoes, we have the summary

standard observations mean deviation wildtype 88 20.784 12.99 transgenic 99 16.546 10.78

Using the conservative 95% confidence interval based on min n ,n 1 = 87 degrees of freedom, we use { 1 2} > qt(0.975,87) [1] 1.987608 to obtain the interval 12.992 10.782 (20.78 16.55) 1.9876 + =(0.744, 7.733) ± r 88 99 Using the the Welch-Satterthwaite equation, we obtain ⌫ = 169.665. The increase in the number of degrees of freedom gives a slightly narrower interval (0.768, 7.710).

16.1.2 For ordinary linear regression, we have given estimates for the slope and the intercept ↵. For data (x1,y1), (x2,y2) ...,(xn,yn), our model is yi = ↵ + xi + ✏i where ✏i are independent N(0,) random variables. Recall that the estimator for the slope

cov(x, y) ˆ(x, y)= var(x) is unbiased.

Exercise 16.9. Show that the variance of ˆ equals 2/(n 1)var(x). 245 Introduction to the Science of Statistics Interval Estimation

If is known, this suggests a z-interval for a -level confidence interval ˆ z(1 )/2 . ± s pn 1 x Generally, is unknown. However, the variance of the residuals,

1 n s2 = (y (ˆ↵ ˆx ))2 (16.2) u n 2 i i i=1 X is an unbiased estimator of 2 and s / has a t distribution with n 2 degrees of freedom. This gives the t-interval u ˆ su tn 2,(1 )/2 . ± s pn 1 x As the formula shows, the margin of error is proportional to the standard deviation of the residuals. It is inversely proportional to the standard deviation of the x measurement. Thus, we can reduce the margin of error by taking a broader set of values for the explanatory variables. For the data on the humerus and femur of the five specimens of Archeopteryx, we have ˆ =1.197. su =1.982, s = 13.2, and t =3.1824, Thus, the 95% confidence interval is 1.197 0.239 or (0.958, 1.436). x 3,0.025 ± 16.1.3 Sample Proportions Example 16.10 (proportions). For n Bernoulli trials with success parameter p, the sample proportion pˆ has p(1 p) mean p and variance . n The parameter p appears in both the mean and in the variance. Thus, we need to make a choice p˜ to replace p in the confidence interval p˜(1 p˜) pˆ z(1 )/2 . (16.3) ± r n One simple choice for p˜ is pˆ. Based on extensive numerical experimentation, one more recent popular choice is x +2 p˜ = n +4 where x is the number of successes. For population proportions, we ask that the mean number of successes np and the mean number of failures n(1 p) each be at least 10. We have this requirement so that a normal random variable is a good approximation to the appropriate binomial random variable.

Example 16.11. For Mendel’s data the F2 generation consisted 428 for the dominant allele green pods and 152 for the recessive allele yellow pods. Thus, the sample proportion of green pod alleles is 428 pˆ = =0.7379. 428 + 152 The confidence interval, using 428 + 2 p˜ = =0.7363 428 + 152 + 4 is 0.7363 0.2637 0.7379 z(1 )/2 · =0.7379 0.0183z(1 )/2 ± r 580 ± For =0.98, z =2.326 and the confidence interval is 0.7379 0.0426 = (0.6953, 0.7805). Note that this 0.01 ± interval contains the predicted value of p =3/4.

246 Introduction to the Science of Statistics Interval Estimation

Example 16.12. For the difference in two proportions p1 and p2 based on n1 and n2 independent trials. We have, for the difference p p , the confidence interval 1 2

pˆ1(1 pˆ1) pˆ2(1 pˆ2) pˆ1 pˆ2 + . ± s n1 n2

Example 16.13 (transformation of a single parameter). If

(✓ˆ`, ✓ˆu) is a level confidence interval for ✓ and g is an increasing function, then

(g(✓ˆ`),g(✓ˆu)) is a level confidence interval for g(✓)

Exercise 16.14. For the example above, find the confidence interval for the yellow pod genotype.

16.1.4 Summary of Standard Confidence Intervals The confidence interval is an extension of the idea of a of the parameter to an interval that is likely to contain the true parameter value. A level confidence interval for a population parameter ✓ is an interval computed from the sample data having probability of producing an interval containing ✓. For an estimate of a population mean or proportion, a level confidence interval often has the form

estimate t⇤ standard error ± ⇥ 1 where t⇤ is the upper 2 critical value for the t distribution with the appropriate number of degrees of freedom. If the number of degrees of freedom is infinite, we use the standard normal distribution to detemine the critical value, usually denoted by z⇤.

The margin of error m = t⇤ standard error decreases if ⇥ , the confidence level, decreases • the standard deviation decreases • n, the number of observations, increases • The procedures for finding the confidence interval are summarized in the table below.

procedure parameter estimate standard error degrees of freedom s one sample µ x¯ pn n 1 2 2 s1 s2 two sample µ1 µ2 x¯1 x¯2 + See (16.1) n1 n2 1 1 pooled two sample µ1 µ2 x¯1 x¯2 sqp + n1 + n2 2 n1 n2 p˜(1qp˜) x+2 one proportion p pˆ , p˜ = n n+4 1 qpˆ1(1 pˆ1) pˆ2(1 pˆ2) two proportion p1 p2 pˆ1 pˆ2 + n1 n2 1 linear regression ˆ =cov(x, y)/var(x) q su n 2 sxpn 1 247 Introduction to the Science of Statistics Interval Estimation

The first confidence interval for µ µ is the two-sample t procedure. If we can assume that the two samples 1 2 have a common standard deviation, then we pool the data to compute sp, the pooled standard deviation. Matched pair procedures use a one sample procedure on the difference in the observed values. For these tests, we need a sample size large enough so that the central limit theorem is a sufficiently good approx- imation. For one population tests for means, n>30 and data not strongly skewed is a good rule of thumb. For two population tests, n>40 may be necessary. For population proportions, we ask that the mean number of successes np and the mean number of failures n(1 p) each be at least 10. For the standard error for ˆ in linear regression, su is defined in (16.2) and sx is the standard deviation of the values of the explanatory variable.

16.1.5 Interpretation of the Confidence Interval

The confidence interval for a parameter ✓ is based on two statistics - ✓ˆ`(x), the lower end of the confidence interval and ✓ˆu(x), the upper end of the confidence interval. As with all statistics, these two statistics cannot be based on the value of the parameter. In addition, these two statistics are determined in advance of having the actual data. The term confidence can be related to the production of confidence intervals. We can think of the situation in which we produce independent confidence intervals repeatedly. Each time, we may either succeed or fail to include the true parameter in the confidence interval. In other words, the inclusion of the parameter value in the confidence interval is a Bernoulli trial with success probability . For example, after having seen these 100 intervals in Figure 5, we can conclude that the lowest and highest intervals are much less likely that 95% of containing the true parameter value. This phenomena can be seen in the presidential polls for the 2012 election. Three days before the election we see the following spread between Mr. Obama and Mr. Romney 0% -1% 0% 1% 5% 0% -5% -1% 1% 1% with the 95% confidence interval having a margin of error 3% based on a sample of size 1000. Because these ⇠ ⇠ values are highly dependent, the values of 5% is less likely to contain the true spread. ± Exercise 16.15. Perform the computations needed to determine the margin of error in the example above. The following example, although never likely to be used in an actual problem, may shed some insight into the difference between confidence and probability.

Example 16.16. Let X1 and X2 be two independent observations from a uniform distribution on the interval [✓ 1,✓+ 1] where ✓ is an unknown parameter. In this case, an observation is greater than ✓ with probability 1/2, and less than ✓ with probability 1/2. Thus, with probability 1/4, both observations are above ✓, • with probability 1/4, both observations are below ✓, and • with probability 1/2, one observation is below ✓ and the other is above. • In the third case alone, the confidence interval contains the parameter value. As a consequence of these considerations, the interval (✓ˆ (X ,X ), ✓ˆ (X ,X )) = (min X ,X , max X ,X ) ` 1 2 u 1 2 { 1 2} { 1 2} is a 50% confidence interval for the parameter. Sometimes, max X ,X min X ,X > 1. Because any subinterval of the interval [✓ 1,✓+ 1] that has { 1 2} { 1 2} length at least 1 must contain ✓, the midpoint of the interval, this confidence interval must contain the parameter value. In other words, sometimes the 50% confidence interval is certain to contain the parameter. Exercise 16.17. For the example above, show that P confidence interval has length > 1 =1/4. { } Hint: Draw the square [✓ 1,✓+ 1] [✓ 1,✓+ 1] and shade the region in which the confidence interval has length ⇥ greater than 1.

248 Introduction to the Science of Statistics Interval Estimation

0.5

0.4

0.3

0.2

0.1

0

−0.1

−0.2

−0.3

−0.4

−0.5 0 10 20 30 40 50 60 70 80 90 100

Figure 16.5: One hundred confidence build from repeatedly simulating 100 standard normal random variables and constructing 95% confidence intervals for the mean value - 0. Note that the 24th interval is entirely below 0 and so does not contain the actual parameter. The 11th, 80th and 91st intervals are entirely above 0 and again do not contain the parameter.

16.1.6 Extensions on the Use of Confidence Intervals Example 16.18 (delta method). For estimating the distribution µ by the sample mean X¯, the delta method provides an alternative for the example above. In this case, the standard deviation of g(X¯) is approximately

g0(µ) | | . pn

We replace µ with X¯ to obtain the confidence interval for g(µ)

g0(X¯) g(X¯) z | | . ± ↵/2 pn

Using the notation for the example of estimating ↵3, the coefficient of volume expansion based on independent length measurements, Y1,Y2,...,Yn measured at temperature T1 of an object having length `0 at temperature T0.

249 Introduction to the Science of Statistics Interval Estimation

¯ 3 3 ¯ 2 Y `0 3Y Y z(1 )/2 `3 T T ± n 0| 1 0| For multiple independent samples, the simple idea using the transformation in the Example 12 no longer works. For example, to determine the confidence interval using X¯1 and X¯2 above, the confidence interval for g(µ1,µ2), the delta methos gives the confidence interval

2 2 2 2 ¯ ¯ @ ¯ ¯ 1 @ ¯ ¯ 2 g(X1, X2) z(1 )/2 g(X1, X2) + g(X1, X2) . ± @x n @y n s✓ ◆ 1 ✓ ◆ 2 A comparable formula gives confidence intervals based on more than two independent samples

Example 16.19. Let’s return to the example of n` and nh measurements y and y of, respectively, the length ` and the height h of a right triangle with the goal of giving the angle

1 h ✓ = g(`, h) = tan ` ✓ ◆ between these two sides. Here are the measurements:

>x [1] 10.224484 10.061800 9.945213 9.982061 9.961353 10.173944 9.952279 9.855147 [9] 9.737811 9.956345 >y [1] 4.989871 5.090002 5.021615 4.864633 5.024388 5.123419 5.033074 4.750892 4.985719 [10] 4.912719 5.027048 5.055755 > mean(x);sd(x) [1] 9.985044 [1] 0.1417969 > mean(y);sd(y) [1] 4.989928 [1] 0.1028745

The angle ✓ is the arctangent, here estimated using the mean and given in radians

>(thetahat<-atan(mean(y)/mean(x))) [1] 0.4634398

Using the delta method, we have estimated the standard deviation of these measurements.

2 2 1 2 ` 2 h ✓ˆ 2 2 h + ` . ⇡ h + ` s n` nh

We estimate this with the sample means and standard deviations

2 s2 1 2 sx 2 y s✓ˆ 2 2 y¯ +¯x =0.0030. ⇡ y¯ +¯x s n` nh

This gives a level z- confidence interval ˆ ✓ z(1 )/2sˆ. ± ✓ For a 95% confidence interval, this is 0.4634 0.0059 = (0.4575, 0.4693) radians or (26.22, 26.89). ± 250 Introduction to the Science of Statistics Interval Estimation

We can extend the Welch and Satterthwaite method to include the delta method to create a t-interval with effective degrees of freedom 2 2 2 2 2 @g(¯x,y¯) s1 + @g(¯x,y¯) s2 @x n1 @y n2 ⌫ = 4 4 4 4 . @g(¯x,⇣y¯) s1 @g(¯x,y¯) ⌘s2 @x n2 (n 1) + @y n2 (n 1) 1· 1 2· 2 We compute to find that ⌫ = 19.4 and then use the t-interval ˆ ✓ t⌫,(1 )/2sˆ. ± ✓

For a 95% confidence, this is sightly larger interval 0.4634 0.0063 = (0.4571, 0.4697) radians or (26.19, 26.91). ± Example 16.20 (maximum likelihood estimation). The Fisher information is the main tool used to set an confidence interval for a maximum likelihood estimation. Two choices are typical. First, we can use the Fisher information In(✓). Of course, the estimator is for the unknown value of the parameter ✓. This give an confidence interval 1 ✓ˆ z . ± ↵/2 nI(✓ˆ) q More recently, the more popular method is to use the observed information based on the observations x = (x1,x2,...,xn). @2 n @2 J(✓ˆ)= log L(✓ˆ x)= log f (x ✓ˆ). @✓2 | @✓2 X i| i=1 X This is the second derivative of the score function evaluated at the maximum likelihood estimator. Then, the confidence interval is 1 ✓ˆ z . ± ↵/2 J(✓ˆ) q Note that E✓J(✓)=nI(✓), the Fisher information for n observations. Thus, by the law of large numbers 1 J(✓) I(✓) as n . n ! !1 If the estimator is consistent and I is continuous at ✓, then 1 J(✓ˆ) I(✓) as n . n ! !1

16.2 The Bootstrap

The confidence regions have been determined using aspects of the distribution of the data. In particular, these regions have often been specified by appealing to the central limit theorem and normal approximations. The notion behind bootstrap techniques begins with the concession that the information about the source of the data is insufficient to perform the analysis to produce the necessary description of the distribution of the estimator. This is particularly true for small data sets or highly skewed data. The strategy is to take the data and treat it as if it were the distribution underlaying the data and to use a protocol to describe the estimator. For the example above, we estimated the angle in a right triangle by estimating ` and h, the lengths two adjacent sides by taking the mean of our measurements and then computing the arctangent of the ratio of these means. Using the delta method, our confidence interval was based on a normal approximation of the estimator. The bootstrap takes another approach. We take the data

x1,x2,...,xn1 ,y1,y2,...,yn2 ,

251 Introduction to the Science of Statistics Interval Estimation

Histogram of angle 1500 1000 Frequency 500 0

26.6 26.8 27.0 27.2

angle

Figure 16.6: Bootstrap distribution of ✓ˆ. the empirical distribution of the measurements of ` and h and act as if it were the actual distribution. The next step is the use the data and randomly create the results of the experiment many times over. In the example, we choose, with replacement n1 measurements from the x data and n2 measurements from the y data. We then compute the bootstrap means x¯b and y¯b and the estimate 1 y¯b ✓ˆ(¯x , y¯ ) = tan . b b x¯ ✓ b ◆ Repeating this many times gives an empirical distribution for the estimator ✓ˆ. This can be accomplish in just a couple lines of R code.

> angle<-rep(0,10000) > for (i in 1:10000){xb<-sample(x,length(x),replace=TRUE); yb<-sample(y,length(y),replace=TRUE);angle[i]<-atan(mean(yb)/mean(xb))*180/pi} > hist(angle)

We can use this bootstrap distribution of ✓ˆ to construct a confidence interval.

> q<-c(0.005,0.01,0.025,0.5,0.975,0.99,0.995) > quantile(angle,q) 0.5% 1% 2.5% 50% 97.5% 99% 99.5% 26.09837 26.14807 26.21860 26.55387 26.86203 26.91486 26.95065

A 95% confidence interval (26.21, 26.86) can be accomplished using the 2.5th as the lower end point and the 97.5th percentile as the upper end point. This confidence interval is very similar to the one obtained using the delta method.

Exercise 16.21. Give the 98% bootstrap confidence interval for the angle in the example above.

252 Introduction to the Science of Statistics Interval Estimation 4 3 2 1 0

0.0 0.2 0.4 0.6 0.8 1.0

x Figure 16.7: Bayesian . The 95% credible interval based on a Beta(9, 3) prior distribution and data consisting of 8 heads and 6 tails. This interval has probability 0.025 both above and below the end of the interval. Because the density is higher for the upper value, an narrow credible interval can be obtained by shifting the values upward so that the densities are equal at the endpoints of the interval and (16.4) is satisfied. 16.3

A Bayesian interval estimate is called a credible interval. Recall that for the Bayesian approach to statistics, both the data and the parameter are random Thus, the interval estimate is a statement about the of the parameter ✓.

P ⇥˜ C(X) X = x = . (16.4) { 2 | } Here ⇥˜ is the random variable having a distribution equal to the ⇡. We have choices in defining this interval. For example, we can choose the narrowest interval, which involves choosing those values of highest posterior density. • choosing the interval in which the probability of being below the interval is as likely as being above it. • We can look at this by examining the two examples given in the Introduction to Estimation. Example 16.22 (coin tosses). In this example, we began with a beta prior distribution. Consequently, the posterior distribution will also be a member of the beta family of distributions. We then flipped a coin n = 14 times with 8 heads. Here, again, is the summary. prior data posterior ↵mean variance heads tails ↵mean 6 6 1/2 1/52 8614 12 7/13 9 3 3/4 3/208 8617 9 17/26 3 9 1/4 3/208 8611 15 11/26 We use the R command qbeta to find the credible interval. For the second case in the table above and with =0.95, we find values that give the lower and upper 2.5% of the posterior distribution. > qbeta(0.025,17,9) [1] 0.4649993 > qbeta(0.975,17,9) [1] 0.8202832

253 Introduction to the Science of Statistics Interval Estimation

This gives a 95% credible interval of (0.4650, 0.8203). This is indicated in the figure above by the two vertical lines. Thus, the area under the density function from the vertical lines outward totals 5%. The narrowest credible interval is (0.4737, 0.8276). At these values, the density equals 0.695. The density is lower for more extreme values and higher between these values. The beta distribution has a probability 0.0306 below the lower value for the credible interval and 0.0194 above the upper value satisfying the criterion (16.4) with =0.95. Example 16.23. For the example having both a normal prior distribution and normal data, we find that we also have a normal posterior distribution. In particular, if the prior is normal, mean ✓0, variance 1/ and our data has sample mean x¯ and each observation has variance 1. The the posterior distribution has mean n ✓ (x)= ✓ + x.¯ 1 + n 0 + n and variance 1/(n + ). Thus the credible interval is 1 ✓1(x) z↵/2 . ± p + n

16.4 Answers to Selected Exercises

16.4. Using R to find upper tail , we find that > qt(0.95,99) [1] 1.660391 > qt(0.99,99) [1] 2.364606 For the 90% confidence interval 79.0 299, 852.4 1.6604 = 299852.4 13.1 or the interval (299839.3, 299865.5). ± p100 ± For the 98% confidence interval 79.0 299, 852.4 2.3646 = 299852.4 18.7 or the interval (299833.7, 299871.1). ± p100 ±

16.8. Let 2 2 2 s1/n1 s2 s1 c = 2 . Then, = c . s2/n2 n2 n1 2 2 Then, substitute for s2/n2 and divide by s1/n1 to obtain 2 2 s2 s2 s2 cs2 1 + 2 1 + 1 2 2 n1 n2 n1 n1 (1 + c) (n 1)(n 1)(1 + c) ⌫ = = = = 1 2 . s4 s4 s4 c2s4 1 c2 2 1⇣ ⌘ 2 ⇣1 ⌘ 1 (n2 1) + (n1 1)c 2 + 2 2 + 2 n 1 + n 1 n (n1 1) n (n2 1) n (n1 1) n (n2 1) 1 2 1· 2· 1· 1· Take a derivative to see that d⌫ ((n 1) + (n 1)c2) 2(1 + c) (1 + c)2 2(n 1)c =(n 1)(n 1) 2 1 · · 1 dc 1 2 ((n 1) + (n 1)c2)2 2 1 ((n 1) + (n 1)c2) (1 + c)(n 1)c = 2(n 1)(n 1)(1 + c) 2 1 1 1 2 ((n 1) + (n 1)c2)2 2 1 (n 1) (n 1)c = 2(n 1)(n 1)(1 + c) 2 1 1 2 ((n 1) + (n 1)c2)2 2 1

254 Introduction to the Science of Statistics Interval Estimation

So the maximum takes place at c =(n 1)/(n 1) with value of ⌫. 2 1 (n 1)(n 1)(1 + (n 1)/(n 1))2 ⌫ = 1 2 2 1 (n 1) + (n 1)((n 1)/(n 1))2 2 1 2 1 (n 1)(n 1)((n 1) + (n 1))2 = 1 2 1 2 (n 1)2(n 1) + (n 1)(n 1)2 1 2 1 2 ((n 1) + (n 1))2 = 1 2 = n + n 2. (n 1) + (n 1) 1 2 1 2 Note that for this value s2 n n /(n 1) 1 = 1 c = 1 1 s2 n n /(n 1) 2 2 2 2 and the variances are nearly equal. Notice that this is a global maximum with

⌫ n 1 as c 0 and s s and ⌫ n 1 as c and s s . ! 1 ! 1 ⌧ 2 ! 2 !1 2 ⌧ 1 The smaller of these two limits is the global minimum. 16.9. Recall that ˆ is an unbiased estimator for , thus E ˆ = , and E [(ˆ )2] is the variance of ˆ. (↵,) (↵,) 1 n n ˆ(x, y) = (x x¯)(y y¯) (x x¯)(x x¯) (n 1)var(x) i i i i i=1 i=1 ! X X 1 n = (x x¯)(y y¯ (x x¯)) (n 1)var(x) i i i i=1 ! X 1 n = (x x¯)((y x ) (¯y x¯)) (n 1)var(x) i i i i=1 ! X 1 n n = (x x¯)(y x ) (x x¯)(¯y x¯) (n 1)var(x) i i i i i=1 i=1 ! X X The second sum is 0. For the first, we use the fact that y x = ↵ + ✏ . Thus, i i i 1 n 1 n Var (ˆ)=Var (x x¯)(↵ + ✏ ) = (x x¯)2Var (↵ + ✏ ) (↵,) (↵,) (n 1)var(x) i i (n 1)2var(x)2 i (↵,) i i=1 ! i=1 X X 1 n 2 = (x x¯)22 = (n 1)2var(x)2 i (n 1)var(x) i=1 X Because the ✏i are independent, we can use the Pythagorean identity that the variance of the sum is the sum of the variances. 16.14. The confidence interval for the proportion yellow pod genes 1 p is (0.2195, 0.3047). The proportion of yellow pod phenotype is (1 p)2 and a 95% confidence interval has as its endpoints the square of these numbers - (0.0482, 0.0928).

16.15. The critical value z0.025 =1.96. For pˆ =0.468 and n = 1500, the number of successes is x = 702. The margin of error is pˆ(1 pˆ) z0.025 =0.025. r n

255 Introduction to the Science of Statistics Interval Estimation

16.17. On the left is the square [✓ 1,✓ + 1] [✓ 1,✓ + 1]. 1 ⇥ For the random variables X1,X2, because they are independent and uniformly distributed over a square of area 4, their joint density is 1/4 on this square. The two diagonal line segments are the graph of x 0.5 | 1 x =1. In the shaded area, the region x x > 1, is precisely the 2| | 1 2| region in which max x ,x min x ,x > 1. Thus, for these { 1 2} { 1 2} 0 values of the random variables, the confidence interval has length greater than 1. The area of each of the shaded triangles is 1/2 1 · · 1=1/2. Thus, the total area of the two triangles, 1, represents a −0.5 probability of 1/4.

16.21. A 98% confidence interval (26.14, 26.91) can be accom- −1 plished using the 1st percentile as the lower end point and the 99th −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 percentile as the upper end point.

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